Atoms: Important Questions

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Very Short Answers [1 Mark]

Ques. Define the meaning of ionization energy. What is the value of ionization energy for a hydrogen atom? (All India 2010)

Ans. The energy needed to knock out an electron from an atom is known as the ionization energy of the atom. For hydrogen atoms, the value of ionization energy is 13.6 eV.

Ques. What is the series of hydrogen spectrum lying in ultraviolet and visible regions?

Ans. In the ultraviolet region lies the Lyman series, whereas in the visible region, the Balmer series is present.

Ques. Define Ionisation energy.

Ans. The energy required to knock out an electron from an atom is called the ionization energy of the atom.

Ques. Determine the maximum number of spectral lines emitted by a hydrogen atom during its third excited state. (CBSE AI 2013C)

Ans. The maximum numbers of spectral lines that can be emitted by a hydrogen atom when it is in the third excited state are six.

Ques. When is the Hα line of the Balmer series in the hydrogen atom's emission spectrum gained? (CBSE Delhi 2013C)

Ans. The Hα line of the Balmer series is gained when an electron jumps to n = 2 level from n=3 level.

Ques. What is the radius of the first excited state of the hydrogen atom, if the radius of its ground level is 5.3 nm?

Ans. It can be calculated as 4 × 5.3 = 21.2 nm 

Ques. In the second orbit, what is Bohr’s quantization condition for the angular momentum of an electron?

Ans. As we all know, L= \(\frac{nh}{2Π} \)
It is given,
n = 2
L = \(\frac{2h}{2Π} \)
∴ L = \(\frac{h}{Π} \)

Ques. Define Excitation energy.

Ans. The amount of energy absorbed by an electron to go from the ground state or lower energy state to a higher energy state is called excitation energy.

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Short Answers [2 Marks]

Ques. Elaborate electron-volt and atomic mass units. Show the calculation of the energy in joule equivalent to the mass of one proton.

Ans. An Electron volt is basically the energy an electron has gained when accelerated through a potential difference of 1 volt.

Atomic mass unit refers to one-twelfth of the mass of one atom of carbon 12.

Energy in Joule Equivalent:

The mass of a proton is 1.67 × 10^-27 kg.

Hence, the energy equivalent of this mass is, E = mc²

= 1.67 × 10^-27 × (3 × 10⁸)2 = 1.5 × 10^-10J

Ques. How to calculate the wavelength of the electron orbiting in the ground state of the hydrogen atom? (CBSE 2018, Delhi 2018)

Ans. The wavelength of an electron in the ground state of a hydrogen atom is given by

E = \(\frac{hc}{Λ} \) or λ = hcE

For ground state

E = – 13.6 eV = 13.6 × 1.6 × 10^-19 J

Henceforth, the wavelength of the electron in the first orbit is, λ= hc/ E

= 6.6×10^-34× 3× 10⁸/13.2×1.6×10^-19 = 0.9×10^-7 J

Ques. What are the kinetic and potential energies of the electron when the ground state energy of the hydrogen atom is −13.6eV?

Ans. Given that the ground state energy of a hydrogen atom, E=−13.6eV which is the total energy of a hydrogen atom. As we know, the kinetic energy is equivalent to the negative of the total energy.

Kinetic energy = −E = − (−13.6) = 13.6eV

The potential energy is equivalent to twice the negative kinetic energy.

Hence, Potential energy = −2× (13.6) = −27.2eV.

Ques. Assuming using a thin sheet of solid hydrogen in place of the gold foil, you are given a chance to repeat the alpha-particle scattering experiment. (Given that hydrogen is solid at temperatures below 14K.) Explain the outcome.

Ans. When a thin solid hydrogen sheet is replaced with a gold foil in an alpha-particle scattering experiment, the scattering angle would not be large enough. It’s because the mass of hydrogen (1.67×10⁻²⁷ kg) is less than that of the mass of incident α-particles (6.64 × 10⁻²⁷kg). Hence, the mass of the target nucleus (Hydrogen) is less than that of the scattering particle. This results in the α-particles not bouncing back if solid hydrogen is used in the α-particle scattering experiment.

Ques. A nucleus makes a transition from one permitted energy level to another level of lower energy.

• Describe the emitted photon belonging to the electromagnetic spectrum.
• Mention the order of its energy in electron volts.
• Write four characteristics of nuclear forces.

Ans.

• Gamma-rays part of the electromagnetic spectrum is where emitted photons belong.
• The order of its energy in electron volts is MeV.
• Four characteristics of nuclear forces are:
• They are independent of charges.
• They are the strongest forces in nature, in their own small range of few fermis.
• They are saturated forces.
• They are short-range forces.

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Long Answer Questions [3 Marks]

Ques. The half-life period of a given radioactive substance is 30 days. Calculate the time for 3/4^th of its original mass to disintegrate.

Ans. Given, Half-life period, T = 30 days

let in time t, 3/4^th part of the original nuclei (N_o) of radioactive substance disintegrate, then

Number of undecayed nuclei, N = (1 – 3/4)N_o = 1/4 N_o

⇒ N/N_o = 1/4

We have N/N_o= (1/2)^t/T

Where T is the half-life period and t is the time in which N nuclei remain undecayed.
⇒ 1/4 = (1/2)^t/30

⇒ (1/2)² = (1/2)^t/30
⇒ t/30 = 2
⇒ t = 60 days
Hence, the time for 3/4^th of the original mass to disintegrate is 60 days.

Ques. Find the quantum number that features the revolution of the earth around the sun in an orbit of radius 1.5×10¹¹m with orbital speed 3×10⁴ m/s with respect to Bohr's model. The mass of the earth = 6.0×10²⁴ kg.

Ans. We know that,

The radius of the earth’s orbit around the Sun, r =1.5×10¹¹ m

The orbital speed of the Earth, v = 3×10⁴ m/s

The mass of the Earth, m = 6.0 × 10²⁴ kg

Based on Bohr's model, the angular momentum is quantized as well as could be given as:

mvr = \(\frac{nh}{2\pi} \)

Where,

• h= Planck's constant =6.62 × 10⁻³⁴ Js
• n= Quantum number

⇒ n = mvr\(\frac{2\pi}{h} \)

⇒ n =2π × 6 × 10²⁴× 3 × 10⁴× 1.5×10¹¹/6.62×10⁻³⁴

∴ n = 25.61×10⁷³ = 2.6×10⁷⁴

Therefore, the quantum number that features the revolution of the earth is found to be 2.6×10⁷⁴.

Ques. What are the radii of the n = 2 and n = 3 orbits given that the radius of the innermost electron orbit of a hydrogen atom is 5.3×10⁻¹¹m?

Ans. Given that the radius of the innermost orbit of a hydrogen atom, r1=5.3×10−11m

Let r₂ be the radius of the orbit at n=2. It is associated with the radius of the innermost orbit as:

r₂= (n)² r₁ = 4×5.3×10⁻¹¹ = 2.12×10⁻¹⁰m

For n=3, we can write the corresponding electron radius as:

r₃= (n)² r₁ = 9×5.3×10⁻¹¹= 4.77×10⁻¹⁰m

Therefore, the radii of an electron for n=2, as well as n=3 orbits, are respectively found to be equivalent to 2.12×10⁻¹⁰ m and 4.77×10⁻¹⁰ m.

Ques. In an atom, two energy levels separate with a difference of 2.3eV. Find the frequency of radiation emitted while the atom transits from the upper level to the lower level.

Ans. It is given that the separation of 2 energy levels in an atom is,

E= 2.3eV= 2.3×1.6×10⁻¹⁹=3.68×10⁻¹⁹J

Let v be the frequency of radiation that is emitted while the atom makes a transition from the upper level to the lower level.

The energy’s relation is E=hv

In which, h = Planck's constant = 6.62×10⁻⁴Js

= v = E/h

When the given values are substituted,

= ν= 3.68×10⁻¹⁹/6.62×10⁻³² = 6.62×10⁻³² = 5.55×10¹⁴ Hz

Therefore, the frequency of the radiation is calculated as 5.55×10¹⁴ Hz.

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Very Long Answers [5 Marks]

Ques. prove that the speed of electrons in the ground state of a hydrogen atom is equivalent to the speed of electrons in the second excited state of a Li⁺⁺ atom.

Ans. The velocity of an electron in the n^th orbit of an atom is given by

v = 2πKze²/nh

Where

• K = coulomb's constant
• z = atomic number
• e = charge of an electron
• n = principle quantum number (position of the orbit)
• h = Planck’s constant

For the speed of an electron in the ground state of a hydrogen atom, we have

z = 1 and n = 1

⇒ v_e = 2πK(1)e²/(1)h = 2πKe²/h

For the speed of an electron in the second excited state of a Li⁺⁺ atom, we have

z = 3 and n = 3

⇒ v_Li = 2πK(3)e²/(3)h = 2πKe²/h

Since, v_e = v_Li

Therefore, it is proved that the speed of electrons in the ground state of a hydrogen atom is equivalent to the speed of electrons in the second excited state of a Li⁺⁺ atom.

Ques. A 12.5 eV electron beam is utilized to bombard gaseous hydrogen at room temperature. Mention the series of wavelengths that will be emitted.

Ans. Given, the energy of electron Beam = 12.5 eV.

The energy of the gaseous hydrogen in its ground state = −13.6eV.

The energy of gaseous hydrogen when bombarded with an electron beam = −13.6 + 12.5eV = −1.1 eV.

For n=3, E= −13.6/9 = − 1.5eV

Henceforth, the electron has jumped from the n=1 level to the n=3 level.

While de-excitation, the electrons jump from n=3 to n=1 level directly. This would form a line of the Lyman series of the hydrogen spectrum. Here is the relation for wave number for the Lyman series,

1/λ= R (1/l²−1/n²)

R= Rydberg constant = 1.097×10⁷m⁻¹

λ = Wavelength of radiation that is emitted by the transition of the electron

For n=3, we can obtain λ as:

= 1/λ= 1.097×10⁷(1/l2−1/32)

= 1/λ= 1.097×10⁷(1−1/9) = 1.097×10⁷×8/9

= λ= 9/ 8×1.097×10⁷ = 102.55nm

Considering the electron jumping from n=2 level to n=1 level, the wavelength of the radiation can be given as:

= 1/λ= 1.097×10⁷(1/l²−1/2²)

= 1λ= 1.097×10⁷(1−1/4) =1.097×10⁷×3/4

∴λ= 4/1.097×10⁷×3 = 121.54nm

When the transition takes place from n=3 level to n=2 level, then the wavelength of the radiation can be mentioned as:

= 1/λ=1.097×10⁷(1/22−1/32)

= 1/λ=1.097×10⁷(1/4−1/9) =1.097×10⁷×5/36

∴λ= 4/ 5×1.097×10⁷ = 656.33nm

Ques. Give answers to the following questions, understanding the difference between Thomson's model and Rutherford's model.

1. What is the average angle of deflection of alpha particles by a thin foil of gold speculated by Thomson's model or much greater, much less, about the same, than that speculated by Rutherford's model?
2. What is the probability of backward scattering (i.e., scattering of alpha particles at angles greater than 90°)?
3. Having other factors fixed, it is established that for small thickness t, the number of alpha particles scattered at moderate angles is proportional to t experimentally. What opinion does this linear dependence on t give?

Ans. 

1. The answer is about the same. The average angle of deflection of alpha particles by a thin gold foil speculated by Thomson's model is about the same size as that of Rutherford's model. Also, this is the reason that the average angle was considered in both these models.
2. The probability of scattering of alpha-particles at angles that are greater than 90° speculated by Thomson's model is much less than that predicted by Rutherford's model.
3. The scattering is due to single collisions mainly. The single collision chance increases linearly with the number of target atoms. As the number of target atoms increases with an increase in thickness, the probability of collision would be based linearly on the thickness of the target.
    

Ques. In the first excited state, the total energy of an electron of the hydrogen atom is about −3.4eV.

1. In the given state, what is the kinetic energy of the electron?
2. In the given state, what is the potential energy of the electron?
3. If the choice of the zero of potential energy is changed, which of the answers would change?

Ans.

1. Given that the total energy of the electron is E=−3.4eV
   The kinetic energy of the electron is equivalent to the negative of the total energy.
   = K.E=−E
   ∴K.E=− (−3.4) = +3.4eV
   Therefore, the kinetic energy of the electron in the given state is concluded as +3.4eV.

2. As we know, the potential energy (U) of the electron is equivalent to the negative of two times its kinetic energy.
   = U=−2K.E
   ∴U=−2×3.4=−6.8eV
   Thus, the potential energy of the electron in the given state is concluded as −6.8eV

100. The potential energy of a system would be based on the reference point considered. The potential energy of the reference point is considered as zero here. When we change the reference point, the value of the potential energy of the system tends to change as well. As we know that the total energy is the sum of kinetic as well as potential energies; the total energy of the system will tend to change.

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